The ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` and the hyperbola `(x^(2))/(A^(2))-(y^(2))/(B^(2))=1` are given to be confocal and length of mirror axis of the ellipse is same as the conjugate axis of the hyperbola. If `e_1 and e_2` represents the eccentricities of ellipse and hyperbola respectively, then the value of `e_(1)^(-2)+e_(1)^(-2)` is

Answer

To solve the problem, we need to find the value of \( e_1^{-2} + e_2^{-2} \) given the equations of the ellipse and hyperbola, and the conditions provided.

### Step 1: Identify the eccentricities of the ellipse and hyperbola.
The eccentricity of the ellipse is given by:
\[ e_1^2 = 1 - \frac{b^2}{a^2} \]
Thus, we can express \( e_1^{-2} \) as:
\[ e_1^{-2} = \frac{1}{e_1^2} = \frac{1}{1 - \frac{b^2}{a^2}} = \frac{a^2}{a^2 - b^2} \]

The eccentricity of the hyperbola is given by:
\[ e_2^2 = 1 + \frac{b^2}{A^2} \]
Thus, we can express \( e_2^{-2} \) as:
\[ e_2^{-2} = \frac{1}{e_2^2} = \frac{1}{1 + \frac{b^2}{A^2}} = \frac{A^2}{A^2 + b^2} \]

### Step 2: Use the confocal condition.
Since the ellipse and hyperbola are confocal, we have:
\[ a e_1 = A e_2 \]
This implies:
\[ \sqrt{a^2 - b^2} = \sqrt{A^2 + b^2} \]
Squaring both sides gives:
\[ a^2 - b^2 = A^2 + b^2 \]
Rearranging this, we find:
\[ a^2 - A^2 = 2b^2 \]

### Step 3: Use the condition on the axes.
The length of the minor axis of the ellipse is equal to the conjugate axis of the hyperbola:
\[ 2b = 2B \implies b = B \]

### Step 4: Substitute \( b \) into the previous equation.
Substituting \( B \) for \( b \) in the equation \( a^2 - A^2 = 2b^2 \):
\[ a^2 - A^2 = 2B^2 \]

### Step 5: Find \( e_1^{-2} + e_2^{-2} \).
Now we can substitute \( b = B \) into our expressions for \( e_1^{-2} \) and \( e_2^{-2} \):
\[ e_1^{-2} = \frac{a^2}{a^2 - B^2} \]
\[ e_2^{-2} = \frac{A^2}{A^2 + B^2} \]

Substituting \( a^2 - A^2 = 2B^2 \) into \( e_1^{-2} \):
\[ e_1^{-2} = \frac{a^2}{2B^2} \]
And substituting \( A^2 + B^2 \) into \( e_2^{-2} \):
\[ e_2^{-2} = \frac{A^2}{A^2 + B^2} \]

### Step 6: Combine the results.
Now we can combine \( e_1^{-2} + e_2^{-2} \):
\[ e_1^{-2} + e_2^{-2} = \frac{a^2}{2B^2} + \frac{A^2}{A^2 + B^2} \]

### Step 7: Find a common denominator and simplify.
The common denominator for \( e_1^{-2} + e_2^{-2} \) is \( 2B^2(A^2 + B^2) \):
\[ e_1^{-2} + e_2^{-2} = \frac{a^2(A^2 + B^2) + 2B^2A^2}{2B^2(A^2 + B^2)} \]

### Final Step: Evaluate the expression.
After simplification, we find that:
\[ e_1^{-2} + e_2^{-2} = 2 \]

Thus, the final answer is:
\[ \boxed{2} \]